Homework 6 - SolutionsMay 8, 2006

Drew Fustin (dafustin@uchicago.edu)Physics 237 - Nuclear and Elementary Particle Physics

Problem 1Low-energy (s-wave) neutrons are scattered from protons, and the distribution of recoil protons is to beobserved and analyzed. Let the neutron scattering angle be θ in the laboratory coordinate system, and letthe incident neutron kinetic energy be Tn. (a) Show that the protons emerge at an angle φ = 90 − θ withrespect to the direction of the incident neutrons. (b) Show that T ′n = Tn cos2 θ and T ′p = Tn sin2 θ, where T ′

signies the energy after scattering. (c) Show that the laboratory and center-of-mass cross sections are relatedby (dσ/dΩ)lab = (4 cos θ) (dσ/dΩ)cm. (d) Given that the scattering is independent of direction in the center-of-mass system, show that (in the laboratory system) dσ/dT ′p = σ/Tn, where σ is the total cross section. Thislatter result shows that the number of recoil protons observed at any particular energy T ′p (0 ≤ T ′p ≤ Tn) isindependent of T ′p. (e) What is the angular distribution of the recoil protons in the laboratory.

Solution(a) First of all, let's assume that the mass of the proton and the mass of the neutron are equal. By theconservation of energy (where a non-relativistic assumption, T = p2/2m, is okay because we are in the low-energy limit), we have

n2 = n′2 + p′2, (1)

where n is the momentum of the incoming neutron, n′ is the momentum of the outgoing neutron, and p′ is themomentum of the outgoing proton (the momentum of the incoming proton p is zero here because we are in thelaboratory frame). In Eq. 1, we notice that all factors of 2m cancel because the masses are equal. Now, by theconservation of momentum in the x-direction, we have

n = n′ cos θ + p′ cosφ, (2)

and from the conservation of momentum in the y-direction, we have

n′ sin θ = p′ sin φ. (3)

Eliminating n′ using Eq. 3 and plugging into Eq. 2 gives

n = p′(

sin φ

tan θ+ cos φ

). (4)

Squaring Eqs. 3 and 4 and plugging them into Eq. 1 gives

p′2(

sin2 φ

tan2 θ+

2 sinφ cosφ

tan θ+ cos2 φ

)= p′2

(sin2 φ

sin2 θ+ 1

),

or

sin2 φ cos2 θ

sin2 θ− sin2 φ

sin2 θ+

2 sin φ cosφ

tan θ= 1− cos2 φ

− sin2 φ sin2 θ

sin2 θ+

2 sin φ cosφ

tan θ= sin2 φ

sin φ cosφ

tan θ= sin2 φ

cot θ = tanφ.

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This implies φ = π/2− θ.(b) Eq. 1 is another form of the conservation of energy, that is

Tn = T ′n + T ′p, (5)

or, after rearranging,

T ′n = Tn

(1− T ′p

Tn

). (6)

Now, the kinetic energy ratio on the right-hand-side of Eq. 6 is nothing more than

T ′pTn

=p′2

n2,

because the masses of the proton and the neutron are approximately equal. Using Eq. 4, this becomes

T ′pTn

=(

sin φ

tan θ+ cos φ

)−2

=tan2 θ

(sinφ + cos φ tan θ)2(7)

Plugging Eq. 7 into Eq. 6, we nd

T ′n = Tn

(1− tan2 θ

(sinφ + cosφ tan θ)2

). (8)

Now, if φ = π/2− θ, then Eq. 8 becomes

T ′n = Tn

(1− tan2 θ(

cos θ + sin2 θ/ cos θ)2

)

= Tn

(1− tan2 θ

(1/ cos θ)2

)

= Tn

(1− sin2 θ

)

= Tn cos2 θ. (9)

Then, Eqs. 5 and 9 together imply T ′p = Tn sin2 θ, as well.(c) Since the cross section doesn't change depending on the reference frame, we simply need to look at the ratioof the solid angles, that is dΩcm/dΩlab. Now, from Problem Set 2, problem 3, we saw that in general

cos θlab =cos θcm + δ

(1 + 2δ cos θcm + δ2)1/2,

where δ = mn/mp. In our case, δ ' 1, and so this reduces to

cos θlab =

√1 + cos θcm

2= cos (θcm/2) . (10)

Taking the dierential of both sides of Eq. 10, we nd

sin θlab dθlab =12

sin (θcm/2) dθcm,

ordθcm

dθlab=

2 sin θlab

sin (θcm/2). (11)

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Now, the ratio of the dierential solid angles is then

dΩcm

dΩlab=

2π sin θcm dθcm

2π sin θlab dθlab

=2 sin θcm sin θlab

sin θlab sin (θcm/2)

=4 sin (θcm/2) cos (θcm/2)

sin (θcm/2)= 4 cos (θcm/2)= 4 cos θlab,

where the last step comes from Eq. 10. Therefore, we nd(

dσ

dΩ

)

lab

= 4 cos θlab

(dσ

dΩ

)

cm

, (12)

which is what we are trying to show, since θ = θlab.(d) Since T ′p = Tn sin2 θ, we see

dT ′p = 2Tn cos θ sin θ dθ

=Tn

πcos θ dΩ. (13)

Therefore,

dσ

dT ′p=

π

Tn cos θ

dσ

dΩ, (14)

which is in the laboratory frame. If we switch to the center-of-mass frame, we can evoke Eq. 12 to nd

dσ

dT ′p=

4π

Tn

(dσ

dΩ

)

cm

. (15)

Now, since the scattering is independent of direction in the center-of-mass system, we know that

σ =∫ (

dσ

dΩ

)

cm

dΩcm

= 4π

(dσ

dΩ

)

cm

,

and so Eq. 15 gives

dσ

dT ′p=

σ

Tn.

(e) Again, since the scattering is independent of direction in the center-of-mass system, we know from Eq. 12that

(dσ

dΩ

)

lab

=σ

πcos θ.

This is the angular distribution in the laboratory.

Problem 2Give the expected shell-model spin and parity assignments for the ground states of (a) 7Li; (b) 11B; (c) 15C;(d) 17F; (e) 31P; (f) 141Pr.

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SolutionBasically, we need to look at which number is odd (either Z, the number of protons, or N , the number ofneutrons) for each nucleus and ll the shell levels starting from the bottom of Fig. 1 and working our way up aseach shell is lled. The nal uncoupled neutron or proton's state determines the spin and parity of the nucleus.The state s corresponds to l = 0, p corresponds to l = 1, d corresponds to l = 2, f corresponds to l = 3, gcorresponds to l = 4, and so on. The parity is given as (−1)l and the spin is given as the last number labelingthe state (e.g. 1/2 or 11/2). We will denote number of particles in a particular state with a superscript afterthe state name. For instance, if there are 3 nucleons in the 1h9/2 state (which has a spin of 9/2 and a parity of(−1)5 = −1), we will write

(1h9/2

)3. Since this state is unlled, it would determine the spin and parity of thenucleus, which we denote as Iπ where I is the spin and π is the parity. This state has spin and parity given by(9/2)−.(a) 7Li has Z = 3 and N = 4, so we are looking at the protons. These ll the states as follows:

(1s1/2

)2 (1p3/2

)1,and so this has a spin-parity of (3/2)−.(b) 11B has Z = 5 and N = 6, so we are again looking at the protons. This has a structure of

(1s1/2

)2 (1p3/2

)3,and so this also has a spin-parity of (3/2)−.(c) 15C has Z = 6 and N = 9, and the neutrons ll with a structure of

(1s1/2

)2 (1p3/2

)4 (1p1/2

)2 (1d5/2

)1 andso this has a spin-parity of (5/2)+.(d) 17F has Z = 9 and N = 8, and the protons ll with a structure of

(1s1/2

)2 (1p3/2

)4 (1p1/2

)2 (1d5/2

)1 andgive a spin-parity of (5/2)+.(e) 31P has Z = 15 and N = 16, and the protons ll with a structure of

(1s1/2

)2 (1p3/2

)4 (1p1/2

)2 (1d5/2

)6 (2s1/2

)1

and give a spin-parity of (1/2)+.(f) 141Pr has Z = 59 and N = 82, and the protons ll with a structure lled up to (we'll just show the lastthree shells)

(1g9/2

)10 (1g7/2

)8 (2d5/2

)1 and give a spin-parity of (5/2)+.

Figure 1: Shell structure of the nucleus

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Problem 3The level scheme of Fig. 1 would lead us to expect Iπ = (11/2)− for the ground state of 203Tl (Z = 81), whilethe observed value is (1/2)+. A similar case occurs in 207Pb (N = 125) and 199Hg (N = 119), where (13/2)+

is expected but (1/2)− is observed. Given that the pairing force increases strongly with l, give the shell-modelcongurations for these nuclei that are consistent with the observed spin-parity assignments.

SolutionThe higher l is, the stronger the pairing force for the nucleons. Therefore, if a state with l = 5 (that is somethingin a state h) is directly above a state with l = 0 (state s), it is reasonable to conclude that nucleons might preferto be coupled in the h state rather than the s state because of the increased pairing force. In this case, eventhough the h state has a higher energy than the s state, the pairing force makes up for that and the h state islled before the s state.

Looking at the last three shells of the nuclei in question, we see that we expect for 203Tl would have aproton structure of

(2d3/2

)4 (3s1/2

)2 (1h11/2

)11 with spin-parity (11/2)−, but because of the pairing force, thelast state gets lled rst and we actually have

(2d3/2

)4 (3s1/2

)1 (1h11/2

)12 which gives a spin-parity of (1/2)+.Similarly for 207Pb whose neutron structure would expected to be

(3p3/2

)4 (3p1/2

)2 (1i13/2

)13 with spin-parity(13/2)+, but it is actually

(3p3/2

)4 (3p1/2

)1 (1i13/2

)14 with spin-parity (1/2)−. 199Hg has an expected neutronstructure of

(3p3/2

)4 (3p1/2

)2 (1i13/2

)7 with spin-parity (13/2)+, but it is actually(3p3/2

)4 (3p1/2

)1 (1i13/2

)8

with spin-parity (1/2)−.

Problem 4Compute the values of the magnetic dipole moments expected from the shell model, and compare with theexperimental values:

Nuclide Iπ µ(exp) (µN)75Ge (1/2)− +0.51087Sr (9/2)+ −1.09391Zr (5/2)+ −1.30447Sc (7/2)− +5.34

147Eu (11/2)− +6.06

SolutionThe magnetic dipole moment of a nucleus is again determined by the last uncoupled nucleon. This time,however, it matters if the last uncoupled nucleon is a proton or a neutron. It also depends on whether the spinis equal to l + 1/2 or l − 1/2. If j = l + 1/2 , the magnetic moment is given by

〈µ〉 =[gl

(j − 1

2

)+

12gs

]µN , (16)

and if j = l − 1/2, the magnetic moment is given by

〈µ〉 =

[gl

j(j + 3

2

)

(j + 1)− 1

21

(j + 1)gs

]µN . (17)

In Eqs. 16 and 17, gl = 1 and gs = 5.586 for protons and gl = 0 and gs = −3.826 for neutrons.75Ge has an odd number of neutrons (N = 43) whose shell structure is

(2p1/2

)1 (1g9/2

)4. Therefore, j = 1/2and l = 1, so we will use the j = l − 1/2 equation, Eq. 17 for neutrons, and nd µ = 1.275µN .

87Sr has an odd number of neutrons (N = 49) whose shell structure is(2p1/2

)2 (1g9/2

)9. Therefore, j = 9/2and l = 4, so we will use the j = l + 1/2 equation, Eq. 16 for neutrons, and get µ = −1.913µN .

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91Zr has an odd number of neutrons (N = 51) whose shell structure is(1g9/2

)10 (1g7/2

)0 (2d5/2

)1. Therefore,j = 5/2 and l = 2, so we will use Eq. 16 for neutrons and get µ = −1.913µN .

47Sc has an odd number of protons (Z = 21) whose shell structure is(1d3/2

)4 (1f7/2

)1. Therefore, j = 7/2and l = 3, so we will use Eq. 16 for protons and get µ = 5.793µN .

Finally, 147Eu has an odd number of protons (Z = 63) whose shell structure apparently ends with(1h11/2

)

to give it a spin-parity (11/2)− with j = 11/2 and l = 5. In this case, we'll use Eq. 16 for protons and getµ = 7.793µN .

All of these predictions are decently o from the observed values, but they are within the ballpark and havethe right sign. Further, they follow the same trend as well. It's within reason that these predictions really aren'tso bad at all.

Problem 5By tabulating the possible m states of three quadrupole (l = 2) phonons, and their symmetrized combinations,show that the permitted resultant states are 0+, 2+, 3+, 4+, and 6+.

SolutionWith three quadrupoles, we have the possibility to get anywhere from µ = +6 to µ = −6 by combining µ1, µ2,and µ3 which have values between −2 and +2. To get µ = +6, we require (µ1, µ2, µ3) = (+2, +2, +2). This isthe only way to get µ = +6, so there is one combination yielding this. We can get µ = +5 in the following ways:(+2, +2, +1), (+2,+1, +2), and (+1, +2, +2). However, these three states are symmetric and therefore combineto form one symmetric wavefunction. Therefore, there is only one way to get µ = +5 as well. To get µ = +4,we can combine in the following ways (ignoring symmetric choices): (+2, +2, 0) and (+2, +1, +1). Therefore,there are two ways to get µ = +4. To get µ = +3, we have: (+2,+1, 0), (+1,+1, +1), and (+2, +2,−1).So, there are three ways to get µ = +3. For µ = +2, we have: (+2, 0, 0), (+1, +1, 0), (+2, +2,−2), and(+2, +1,−1). So, there are four ways to get µ = +2. For µ = +1, we have: (+1, 0, 0), (+2,−1, 0), (+1, +1,−1),and (+1,+2,−2). Therefore, there are four ways to get µ = +1. For µ = 0, we have: (0, 0, 0), (+2,−2, 0),(+1,−1, 0), (+2,−1,−1), and (−2, +1,+1). So, there are ve ways to get µ = 0. Each negative µ state wouldhave the same number of combinations as its positive µ counterpart, which is obvious by just replacing everyplus sign with a minus sign.

To summarize, we have one way to get µ = +6, one way to get µ = +5, two ways to get µ = +4, three waysto get µ = +3, four ways to get µ = +2, four ways to get µ = +1, and ve ways to get µ = 0. A state with l = 6has the following allowed µ values: µ = +6, +5, +4,+3, +2, +1, 0,−1,−2,−3,−4,−5,−6. This takes care of theone way to get both µ = +6 and µ = +5. To get another µ = +4 state, we need to add an l = 4 state withallowed µ values of µ = +4, +3,+2, +1, 0,−1,−2,−3,−4. Now, between l = 6 and l = 4, we have two µ = +3states, but we need one more, so we add a l = 3 state with allowed µ values µ = +3,+2, +1, 0,−1,−2,−3. Now,we have three µ = +2 states, so we need to add a l = 2 state with allowed µ values µ = +2, +1, 0,−1,−2. Thisgives four µ = +1 states, so we are good for this value. Finally, to get the extra µ = 0 state we don't alreadyhave, we add a l = 0 state. Therefore, to get the desired structure, we have

l = 6 µ = +6,+5, +4, +3,+2, +1, 0,−1,−2,−3,−4,−5,−6l = 4 µ = +4,+3, +2, +1, 0,−1,−2,−3,−4l = 3 µ = +3,+2, +1, 0,−1,−2,−3l = 2 µ = +2,+1, 0,−1,−2l = 0 µ = 0,

which gives us the right number of each µ state. Since we are working with quadrupoles with λ = 2, the parityof each added state is positive. So the permitted resultant states are given by 0+, 2+, 3+, 4+, and 6+.

Problem 6The levels of 174Hf show two similar rotational bands, with energies given as follows (in MeV):

6

E(0+) E(2+) E(4+) E(6+) E(8+) E(10+) E(12+)Band 1 0 0.091 0.297 0.608 1.010 1.486 2.021Band 2 0.827 0.900 1.063 1.307 1.630 2.026 2.489

Compare the moments of inertia of these two bands, and comment on any dierence.

SolutionThe energy of a particular state with angular momentum l is given as a function of the moment of inertia I as

E =~2

2Il (l + 1) . (18)

This tells us, for instance, that the energy of the l = 0 state is 0, as seen in Band 1. Band 2 has a non-zerovalue for the energy here, and so to use Eq. 18 to nd the moment of inertia, we should subtract this energy(0.827 MeV) from each of the energies of Band 2. This gives corrected values of

E'(0+) E'(2+) E'(4+) E'(6+) E'(8+) E'(10+) E'(12+)Band 1 0 0.091 0.297 0.608 1.010 1.486 2.021Band 2 0 0.073 0.236 0.480 0.803 1.199 1.662

where the prime on the energy just implies that these refer to the corrected values for Band 2. Taking the ratioof Eq. 18 for Band 1 to that of Band 2 for the dierent energies gives a ratio of the moments of inertia (called I1

for Band 1 and I2 for Band 2). For the 2+ state, we nd I2/I1 = 1.247. For the 4+ state, we nd I2/I1 = 1.258.For the 6+ state, we nd I2/I1 = 1.267. For the 8+ state, we nd I2/I1 = 1.258. For the 10+ state, we ndI2/I1 = 1.239. Finally, for the 12+ state, we nd I2/I1 = 1.216. The last two states state to deviate from theearlier results because for higher l values, we start to see an eect called centrifugal stretching. This eect isdescribed in the text (Krane section 11.13). Overall, discounting the higher l results, we nd a ratio of momentsof inertia of around I2/I1 ' 1.26.

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